Question

Let $P= \left\{1, 2, 3, 4, 8, 16, .......1024\right\}$. Consider all possible positive differences of elements of $P$.Let $M$ is the sum of all these differences and $N$ be the sum of the digits of $M$ then

• A. $N$ is divisible by $3$ or $9$ or product of $3$ and $9$
• B. Number of prime factors of $N$ are $4$
• C. Sum of divisors of $N$ is $40$
• D. Number of divisors of $N$ are $4$

Answer 1

Answer: (A), (C), (D)

Given $P= \left\{2^{0}, 2^{1},2^{2} .....2^{10}\right\}$
$\therefore M =$Sum of all difference $= \sum\limits^{10}_{s = 1} \sum\limits^{s -1}_{r =0}\left(2^{s} -2^{r}\right)$
$\therefore M = \sum\limits^{10}_{s = 1} \left[\left(2^{s} -2^{0}\right) +\left(2^{s} -2\right) +\left(2^{s} -2^{2}\right)+ ....+ \left(2^{s} -2^{s -1}\right)\right]$
$= \sum\limits^{10}_{s = 1}\left(2^{s} +2^{s} +....\left(s\, times\right)\right)-\left(2^{0} +2^{1}+......+2^{s-1}\right)]$
$= \sum\limits^{10}_{s = 1}\left[s2^{s}-2^{0} \frac{\left(2^{s}-1\right)}{2 -1}\right]$
$= \sum\limits^{10}_{s = 1}\left[\left(s-1\right)2^{s} +1\right]$
$= \sum\limits^{10}_{s = 1}\left(s -1\right)2^{s} +\sum\limits^{10}_{s = 1}1$
$= 0 + 4\left(1 + 2\cdot2 + 3\cdot2^{2} + .... + 9\cdot2^{8}\right) + 10$
$= 4 + 8\cdot2^{11} + 10$ (using A.GP.)
$= 14 + 2^{14} = 14 + \left(1024\right)16 = 16398$
$\therefore N =$ sum of digits of $M = 1 + 6 + 3 + 9 + 8 = 27$
As $N= 27$, which is divisible by $1, 3, 9$ and $27$ thus number of divisors of $N$ are $4$ and their sum is $1 + 3 + 9 + 27 = 40$